Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. [2023]

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Solution

Q.
Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point $P$ is at equidistant from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnitude of magnetic field at $P$ is ________ $\times 10^{- 6} T$ . (Given: $\sqrt{2} = 1.4$ ) [2023]

Ans.
(68)

$Magnetic fields due to both wires will be perpendicular to each other.$

$B_{1} = \frac{μ_{0} i_{1}}{2 π d}, B_{2} = \frac{μ_{0} i_{2}}{2 π d}$

$B_{net} = \sqrt{B_{1}^{2} + B_{2}^{2}} = \frac{μ_{0}}{2 π d} \sqrt{i_{1}^{2} + i_{2}^{2}}$

$\Rightarrow \frac{4 π \times 10^{- 7}}{2 π \times (\frac{7}{\sqrt{2}}) \times 10^{- 2}} \times \sqrt{8^{2} + 15^{2}} (d = \frac{7}{\sqrt{2}} cm)$

$= 68 \times 10^{- 6} T$

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